Consider the following definite integral
$$A = \int_{0}^{3} \left[- \frac {x^2}{4} - 8\right] \,dx$$
Calculate the Riemann sum that approximates the value of A as a closed-form formula in n (i.e. remove the Sigma using the necessary formulas). Use right-handed endpoints.
Next, find the actual value of A by taking a limit of your formula.
No idea where to start.
If you split your interval $[a,b]$ into $n$ even sub-intervals, then we know that the width of each sub-interval is $\Delta x = \dfrac{b-a}{n}$. Let $a=x_0<x_1<\ldots<x_n=b$ evenly partition the interval $[a,b]$ (by evenly partition, I mean $x_{i+1}-x_i=\Delta x$). Then it follows that $x_1=a+\Delta x$, $x_2=a+2\Delta x$, and hence $x_i=a+i\Delta x$ for $1\leq i\leq n$. Since we're after a right Riemann sum, we take the right endpoint to be $x_k^{\ast}=x_k=a+k\Delta$ (see remark at bottom). We can now see that the area of each rectangle is $f(x_k^{\ast})\Delta x$. Thus the definite integral is given by the following right Riemann sum: $$\int_a^b f(x)\,dx =\lim_{n\to\infty}\sum_{k=1}^nf(x_k^{\ast})\Delta x = \lim_{n\to\infty}\frac{b-a}{n}\sum_{k=1}^n f\left(a+\frac{k}{n}(b-a)\right)$$
In the particular problem that you have, we can easily see that $f(x)=-\frac{1}{4}x^2-8$ and that we're computing the integral over the interval $[0,3]$. Hence $\Delta x=\dfrac{3}{n}$ and thus your first step is to find the closed form of the partial sum $$S_n = \sum_{k=1}^n f(x_k^{\ast})\Delta x = \frac{3}{n}\sum_{k=1}^n\left( -\frac{1}{4}\left(\frac{3k}{n}\right)^2 - 8\right)=\ldots$$ I'll leave the rest of the simplification of this sum to you.
Once you finally have the closed form of the above partial sum, take the limit as $n\to\infty$ to get the value of the definite integral you were originally given.
Remark: Depending on the type of Riemann sum one wants to evaluate, $x_k^{\ast}$ will be one of three things:
Right endpoint RS: $x_k^{\ast}=x_k,\, 1\leq k\leq n$
Left endpoint RS: $x_k^{\ast}=x_{k-1},\, 1\leq k\leq n$
Midpoint RS: $x_k^{\ast} = \frac{1}{2}\left(x_{k-1}+x_k\right),\, 1\leq k\leq n$